In a box of $10$ electric bulbs, two are defective. Two bulbs are selected at random one after the other from the box. The first bulb after selection being put back in the box before making the second selection. The probability that both the bulbs are without defect is

The correct option is B

$\frac{16}{25}$

Explanation for the correct option.

Find the probability.

There are $10$ electric bulbs and $2$ are defective. So there are $8$ bulbs without defect.

Probability of getting a defective bulb is:

$\begin{array}{rcl}p& =& \frac{2}{10}\\ & =& \frac{1}{5}\end{array}$

So, probability of getting a bulb without defect is:

$\begin{array}{rcl}q& =& 1-p\\ & =& 1-\frac{1}{5}\\ & =& \frac{4}{5}\end{array}$

Now, let $X$ be the number of defective balls out of $2$ trials. Then the binomial distribution is given as:

$\begin{array}{rcl}\mathit{P}\left(X=x\right)& \mathbf{=}& {}^{\mathbf{n}}\mathit{C}_{\mathbf{x}}{\mathit{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}{\mathit{p}}^{\mathbf{x}}\\ & =& {}^{2}C_x{\left(\frac{4}{5}\right)}^{2-x}{\left(\frac{1}{5}\right)}^x\left[n=2,q=\frac{4}{5},p=\frac{1}{5}\right]\end{array}$

So to find the probability that none of the bulbs is defective substitute $0$ for $x$.

$\begin{array}{rcl}P\left(X=0\right)& =& {}^{2}C_0{\left(\frac{4}{5}\right)}^{2-0}{\left(\frac{1}{5}\right)}^0\\ & =& 1\times \frac{16}{25}\times 1\left[{}^{n}C_0=1,a^0=1\right]\\ & =& \frac{16}{25}\end{array}$

So the probability that both the bulbs are without defect is $\frac{16}{25}$.

Hence, the correct option is B.